Let reaction at support A and B be, RA and RB
First find the support reaction.
For finding the support reaction, convert UDL in to point load and equal to 2 X 2 = 4KN, acting at mid point of UDL i.e. 3m from point A. For that,
∑V = 0
RA + RB – 1 – 4 – 1 = 0,
RA + RB = 6 ...(1)
Taking moment about point A,
∑MA = 0
1 X 1 + 4 X 3 + 1 X 5 – RB X 6 = 0
RB = 3KN
From equation (1), RA = 3KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 5-section line, which break the
load RA and 1KN (Between Point A and C),
1KN and starting of UDL (Between Point C and D),
end point of UDL and 1KN (Between Point E and F) and
1KN and RB (Between Point F and B)
Let Distance of section 1-1 from point A is X1
Distance of section 2-2 from point A is X2
Distance of section 3-3 from point A is X3
Distance of section 4-4 from point A is X4
Distance of section 5-5 from point A is X5
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RA
SF1–1 = 3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFA = SFC = 3KN ...(4)
Consider section 2-2
Forces on left of section 2-2 is RA & 1KN
SF2–2 = 3 – 1 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFC = SFD = 2KN ...(5)
Consider section 3-3 Forces on left of section 3-3 is RA, 1KN and UDL (from point D to the section line i.e. UDL on total distance of (X3 - 2)
SF3–3 = 3 - 1 - 2(X 3 - 2) = 6 - 2 X 3 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear. Inclined linear means value of *.F. at both nearest point of the section is varies with X3 = 2 to
X3 = 4
At X3 = 2
SFD = 2 ...(6)
At X3 = 4 SFE = –2 ...(7)
i.e. inclined line 2 to -2
